Relative scalar: Difference between revisions

Revision as of 20:06, 30 August 2022

Revision as of 20:26, 30 August 2022

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Suppose that one wishes to ''integrate'' these functions over "the room", which will be denoted by <math>D</math>. (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region <math>D</math> is given in cylindrical coordinates as <math>r</math> from <math>[0,2]</math>, <math>t</math> from <math>[0,\pi/2]</math> and <math>h</math> from <math>[0,2]</math> (that is, the "room" is a quarter slice of a cylinder of radius and height 2).

The integral of <math>f</math> over the region <math>D</math> is{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.</math>

The value of the integral of <math>\bar{f}</math> over the same region is{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 12 + 10 \pi.</math>

They are not equal. The integral of temperature is not independent of the coordinate

system used. It is non-physical in that sense, hence "strange". Note that if the integral of <math>\bar{f}</math> included a factor of the Jacobian (which is just <math>r</math>),

we get{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi,</math>

which ''is'' equal to the original integral but it is not however the integral of ''temperature'' because

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<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 16 + 10 \pi.</math>

They are equal. The integral of mass ''density'' gives total mass which is a coordinate-independent concept.

Note that if the integral of <math>\bar{f}</math> also included a factor of the Jacobian like before, we get{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 24 + 40 \pi /3,</math>

which is not equal to the previous case.